Arches 100191

Fig. 111.

Stress at B = 19250/336 + 6. 19250.13/336.28 = + 217 pounds.

And

Stress at A = 19250/336 - 6. 19250/336.28 = - 102 pounds.

The arch, therefore, cannot safely carry such heavy loads. The pier we shall naturally expect to find still more unsafe, and in effect have, remembering that joint D C is 28" wide, therefore, area 336 square inches, and as K1 distant 54" from centre of joint, and p = o h = 21750 pounds.

Thrust on central pier.

Stress at D = 21750/336 + 6. 21750.54/336.28 = + 813 pounds.

and

Stress at C = 21750/336 - 6. 21750.54/336.28 = - 684 pounds.

The construction, therefore, must be radically changed, if the loads cannot be altered. If the arches are needed as ornamental features, they should be constructed to carry their own weight only, and iron-work overhead should carry the loads, and bear either nearer to, or directly over, the piers, as farther trials and calculations might call for. If this is done the wall To avoid should be left hollow under all but the ends of ironcracks, work until it gets its "set"; that is, until it has taken its full load and deflection; and then the wall should be pointed with soft "putty" mortar.

Relief through iron-work.

Example VII. The foundations of a building rest on brick piers 6' wide and 18' apart. The piers are joined by 32" brick inverted arches and tied together 8' above the spring of the arch. Piers and arches are 3' thick. Load on central piers is 72 tons; on end pier 60 tons. Is this construction safe?

We will first examine the inner or left pier. The pier being 3' thick and the load 72 tons, each 1' of tlnckness will, of course, carry 72/3 = 24 tons. The width from centre to centre of piers is just 24', so that each running foot of wall under arch will receive a pressure of one ton. Now all we need to do is to imagine this pressure as the load on the arch. We can either draw the arch upside down with a load of one ton per foot, or we can make the drawing with the arch in correct position and the weights pressing upward, as shown in Fig. 112. We divide the load into five equal slices, each about 2' wide, therefore = 2 tons each.

Inverted arches.

Inverted arches

Fig. 112.

We make ab = 2 tons; b c = 2 tons, etc., and find the horizontal pressures g1 h1 g2 h1, etc., same as before. Again, g5 h5 is the largest, measuring 13 units or tons. Now make o a = 13 tons, and draw o b, o c, etc. Construct the line a 1 K5 or curve of pressure same as before. Joint L M is evidently the most strained one. We find M K5 measures 11", and as the arch is 32" (= M L), of course, K5 is 5" from the centre of joint. The area of joint is a = 32. 12 = 384. We scale of, the pressure at K5, and find that it measures 16 1/2 units, or 33000 pounds, therefore

Strength of arch.

Stress at M = 33000/384 + 6. 33000.5/384.32 = + 168 pounds

Stress at L = 33000/384 - 6. 33000.5/384.32 = + 6 pounds.

The arch, therefore, is perfectly safe. Of course, the left pier is safe, for being an inner pier the resistance K6 x of the adjoining arch to the left will just counter-balance the thrust of our arch, or K5 x. But at the end (right) pier this is different. We, of course, proportion the length of foundation A C, to get same pressure as on rest of wall under arch. The end pier carries 60 tons, or 60/3 = 20 tons per foot thick. The pressure per running foot on wall under arch we found to be one ton, therefore A C should be 20 feet long. The half-arch will take the pressure of 10 feet (from A to B) or 10 tons, and the balance (10 tons) will come on B C. This will act through its central axis G H, which, at the arch skew-back, will be half-way between the end of arch J and outside pier-line I F. This will, of course, deflect somewhat the abutment (or last pressure) line K7 H of the arch. At any convenient place draw a,/, vertically equal 10 tons, and a1 o1, horizontally equal 13 tons, the already known horizontal pressure. Then K7 H is, of course, parallel with and equal olfl. Now make f1 g1 =

10 tons and draw o1 gt. Now on the pier draw H K parallel with o1 g1

H being the point of intersection of G H and K7 H. As the pier is tied back 8' above the arch, we take our joint-line at E F, being 8' above D. We find, by scale, that K is 58" from the centre of E F, the latter being 72" wide. The area is a = 72. 12 = 864. And as o1 g1, scales 24 units, the pressure is, of course, 24 tons or 48000 pounds,

Central pier.

Thrust on end pier.

therefore:

Stress at F = 48000/864 + 6. 48000.58/864.72 = + 315 pounds.

and

Stress at E = 48000/864 - 6. 48000.58/864.72 = - 207 pounds.

There is, therefore, no doubt of the insecurity of the end pier. Two courses for safety are now open. Either we can build a buttress sufficiently heavy to resist the thrust of the end arch, or we can tie the pier back. The former case is easily calculated; we simply include the mass of the buttress in the resistance and shift the axis G H to the centre of gravity of the area of pier and buttress up to joint E F. Of course, the buttress should be carried up to the joint-line E F, but it can taper away from there. If we tie back with iron we need sufficient area to resist a tension equal to the horizontal pressure a o, which in this case is 13 tons or 26000 pounds per foot thick of wall. As the wall is 3' thick, the total horizontal thrust is 3. 26000 = 78000 pounds. The tensional stress of wrought-iron being 12000 pounds per square inch, we need 7800/12000= 6 1/2 square inches area, or, say, two wrought-iron straps, 4" x 7/8", one each side of pier. By this method the inner or left arch becomes practically the end arch. For the last two piers and the right arch become one solid mass; and not only is their entire weight thrown against the second or inner arch, but the centre of gravity of the whole mass shifts to the centre line of end arch, or in our case 0' inside of the end pier; so that there is no possible doubt of the strength of the abutment. There is one element of weakness, however, in the small bearing the pier has on the skew-back of the arch, the danger being of the pier cracking upwards and settling past and under the arch. This can be avoided, as already explained, by building a large bond-stone across the entire pier, forming skew-backs for the arch to bear against.